A solution of ferrous sulfate (FeSO4) is titrated with a solution of potassium permanganate in sulfuric acid. The balanced half reactions are as follows:

10FeSO4 + 5 H2SO4 --> 5 Fe2(SO4)3 + 10 H ^+ +10e^-
and 2 KMnO4 + # H2SO4 +10e^- --> 2 MnO4 + K2SO4 + 8 H2O.

The gram-formula mass of FeSO4 is 151.91 g/mol, and that of KMnO4 is 158.04 g/mol.

1. What is the gram-equavalent mass of FeSo4 for this reaction?

2. What is the gram-equavalent mass of KMnO4 for this reaction?

3. If 5.35 g of KMnO4 are dissolved in a liter of solution, what is the normality of the solution?

4. If 65.34 mL of 0.1000 N KMnO4 are required to titrate 45.28 mL of FeSO4 solution, what is the normality of the solution?

5. Carbon monoxide is detected in air samples by its reaction with diiodine pentoxide to form carbon dioxide and elemental iodine: CO + I2O5 --> CO2 + I2. Balance the equation.

These questions really scare me!!! I have just started this new chapter in my chemistry lessons and I don't get it at all!!!PLEASE HELP!!

I understand that these questions may seem overwhelming at first, but don't worry! I'll guide you through each step and explain how to find the answers.

1. To find the gram-equivalent mass of FeSO4, we need to determine the molar mass of FeSO4 and then divide it by the number of equivalents in the balanced equation.

The molar mass of FeSO4 is given as 151.91 g/mol. The balanced equation shows that 10 moles of FeSO4 react with 10 equivalents of electrons.

Therefore, the gram-equivalent mass of FeSO4 = (151.91 g/mol) / 10 equivalents
= 15.19 g/equivalent

2. Similarly, to find the gram-equivalent mass of KMnO4, we need to determine its molar mass and divide it by the number of equivalents in the balanced equation.

The molar mass of KMnO4 is given as 158.04 g/mol. The balanced equation shows that 10 equivalents of electrons are involved in the reaction.

Therefore, the gram-equivalent mass of KMnO4 = (158.04 g/mol) / 10 equivalents
= 15.804 g/equivalent

3. The normality of a solution is the number of equivalents of solute present in one liter of the solution. To find the normality of the KMnO4 solution, we need to know the number of equivalents of KMnO4 present.

From the balanced equation, we can see that 2 moles of KMnO4 react with 10 equivalents of electrons.

Therefore, the number of equivalents of KMnO4 = (2 * 5.35 g) / (158.04 g/equivalent)
= 0.0677 equivalents

The normality is the number of equivalents divided by the volume in liters. Since the volume is given as 1 liter, the normality of the KMnO4 solution is 0.0677 N.

4. In this question, we are given the volume and normality of KMnO4 solution used to titrate the FeSO4 solution. We need to calculate the normality of the FeSO4 solution.

We can use the formula: N1V1 = N2V2, where N1 and V1 are the normality and volume of the KMnO4 solution, and N2 and V2 are the normality and volume of the FeSO4 solution.

N1 = 0.1000 N (given)
V1 = 65.34 mL = 0.06534 L (convert to liters)
V2 = 45.28 mL = 0.04528 L (convert to liters)

By substituting these values into the formula, we get:
0.1000 N * 0.06534 L = N2 * 0.04528 L

Solving for N2, we find: N2 = (0.1000 N * 0.06534 L) / 0.04528 L
= 0.1443 N

Therefore, the normality of the FeSO4 solution is 0.1443 N.

5. To balance the equation CO + I2O5 -> CO2 + I2, we need to make sure that the same number of atoms of each element is present on both sides of the equation.

Initially, the equation is unbalanced because there is one carbon (C) atom on the left side and one carbon dioxide (CO2) molecule on the right side. We need two CO2 molecules to balance the carbon atoms.

The balanced equation is: CO + I2O5 -> 2CO2 + I2

I2O5 has five oxygen (O) atoms, so we need to have five oxygen atoms on the right side as well. CO2 already has two oxygen atoms, so we just need three more. This can be achieved by adding three oxygen (O2) molecules to the right side.

The final balanced equation is: CO + I2O5 -> 2CO2 + I2 + 3O2

I hope this helps you understand the steps involved in solving these questions. If you have any further questions, feel free to ask!